Simpson's proof of the Gandy Kreisel Tait theorem in $\textbf{ATR}_0$

Question

I believe there is an error in Simpson's book "Subsystems of Second Order Arithmetic". Theorem VIII.6.4 states: $\textbf{ATR}_0$ proves that any $Y$ such that $Y_i=\{(n,i)\in y:n\in\mathbb{N}\}\not\leq_H X$ and if $\exists W\varphi(W,X)$ where $\varphi$ is a $\Sigma^1_1$ formula then $\exists W\;(\varphi(W,X)\wedge \forall i,j W_i\neq Y_j)$.

Proof: By Kleene normal form we have that $\varphi(W,X)$ is equivalent to $\exists f \;\forall n\; \theta(W[n],f[n],X)$. We define $\theta^*(\sigma,\tau, X)$ to be: \begin{equation} \exists W,f\; (W[|\sigma|]=\sigma\wedge f[|\tau|]=\tau \wedge \forall n \;\theta(W[n],f[n],X) \end{equation} which informally states that $(\sigma,\tau)$ are an extendible element of the tree given by the Kleene normal form. By $f[n]$ we mean the sequence given by restriction of $f$ to $[0,n-1]$. We have that the following formulas can be expressed by a $\Sigma^1_1$ formula over $\Sigma^1_1$ axiom of choice: for every $X$ computable well order $\alpha$ the set $X^{(\alpha)}$ exists, $\forall i(Y_i\not\leq_H X)$, and $\exists W\varphi(W,X)$.

By a previous lemma one can construct a coded $\omega$ model of $\textbf{ACA}_0$ in which those propositions listed before hold. We have that $M$ satisfies $\theta(\emptyset,\emptyset,X)$ and we wish to show that if $ M \vDash \theta^* (\sigma,\tau,X)$ then for all $i,j\in \mathbb{N}$ there exists $\sigma'$ that is an extension of $\sigma$ and $M\vDash \theta^* (\sigma',\tau,X)\wedge (Y)_i[|\sigma'|]\neq (\sigma')_j$. If this does not hold then there are $i,j\in\mathbb{N}$ such that for all $m\in \mathbb{N}$ and $k\leq 1$ we have: \begin{equation} (Y)_i(m)=k\leftrightarrow M\vDash \exists \sigma'\supseteq\sigma (\theta^*(\sigma',\tau,X)\wedge (\sigma')_j(m)=k) \end{equation}

Simpson then states that $M$ would then satisfy $(Y)_i$ being $\Delta^1_1$ and therefore $Y_i\leq_H X$. This however is not clear to me, what is clear to me is that $Y_i\leq_H X$ holds outside of $M$ since satisfiabiliy for coded $\omega$-models is $\Delta^1_1$ definable in $\textbf{ACA}_0$. It seems to me that $(Y)_i\leq_H X$ outside of $M$ is also sufficient to get a contradiction. This means that I do not need $M$ to also satisfy $\forall i\;((Y)_i\not\leq_H X)$ and that for every $\alpha$ the set $X^{(\alpha)}$ exists, since I derive the contradiction outside of $M$ where these proposition already hold.

The proof after is straightforward. We construct recursively two increasing chains of sequence of $\sigma_n$ and $\tau_n$ such that $\theta^* (\sigma_n,\tau_n,X)$ and $n=(i,j) $ then $(Y)_i [|\sigma_n|] \neq (\sigma_n)_j$ . Setting $W=\bigcup_{n\in\mathbb{N}}\sigma_n$ gives us the wanted set which concludes the proof.

Is it necessary to show that $(Y)_i$ is $\Delta^1_1$ for the proof to work and is there a mistake in Simpson's proof?

Answer 1

Answering this question now that I have had a better look at Simpson's book. In general, the model $M$ will not be $X$ computable, nor $\Delta_1$ relative to $X$. This is because we do not have any control over the complexity for the witness of $\exists W\,\varphi(W,X)$. So in particular in the argument laid out in the question we would only get $Y_i$ being $\Delta^1_1$ relative to $M$, which does not yield the desired contradiction. I will sketch the proof as I understand it now.

The model $M$ must satisfy the following condition $$\forall a\in \mathbb{N}(\mathcal{O}(a,X)\rightarrow H_a^X\text{ exists })$$ This is to ensure that at least in $M$ the statement $\forall i\in\mathbb{N}\, Y_i\nleq_H X$ makes sense. Note, this does not ensure by any means that in $M$ the right choice is made, namely, in $M$ we may have a pseudohierarchy for $a\notin \mathcal{O}^X$ instead of a descending sequence in the linear order computed by $a$. However, $$(M\vDash Y_i\nleq_H X)\rightarrow Y_i\nleq_H X$$ Furthermore, this condition also implies, by a formalization of the Kleene Souslin theorem ( Theorem VIII.3.19), that if $Z$ is $\Delta^1_1$ relative to $X$ in $M$ then $M\vDash Z\leq_H X$.

Revisiting the proof, we would like to show $$M\vDash \forall \sigma\, i\, j\, (\theta^*(\sigma\,\tau, X)\rightarrow \exists \rho \supseteq \sigma\,( \theta^*(\rho,\tau, X)\wedge Y_i\restriction_{\leq |\rho_j|}\neq \rho_j))$$ If this were not the case then there exists $\sigma, i, $ and $j$ such that $$M\vDash \forall \rho\supseteq \sigma (\theta^*(\rho,\tau,X)\rightarrow Y_i\restriction_{\leq |\rho_j|}= \rho_j)$$ this provides, within $M$ a $\Pi^1_1$ definition for $Y_i$ relative to $X$. As Simpson notes, this is equivalent, since $\sigma$ must be extendible, to $$ \forall m\in \mathbb{N}\, \forall k\leq 1 \;((Y)_i(m)=k\leftrightarrow M\vDash \exists \rho\supseteq \sigma \,(\rho_j(m)=k\wedge \theta^*(\rho,\tau,X))) $$ Since $M$ is an $\omega$-model, this is equivalent to $$ M\vDash\forall m\in \mathbb{N}\, \forall k\leq 1\; ((Y)_i(m)=k\leftrightarrow \exists \rho\supseteq \sigma \,(\rho_j(m)=k\wedge \theta^*(\rho,\tau,X))) $$ which provides a $\Sigma^1_1$ definition of $Y_i$ relative to $X$. So $$M\vDash Y_i\text{ is } \Delta^1_1\text{ relative to } X$$ By the Kleene Souslin theorem, which as we pointed out holds in $M$, it follows that $M\vDash Y_i\leq_H X$. The rest of the proof is in the answer.